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Goddamn trig problem

  1. Jul 1, 2005 #1
    On my math final, there was this bastard of a trig problem that I simply couldn't solve. I knew the equations to use, how to solve the problem, but the answers just didn't work...

    Anyway, the question was this:
    Find all positive values of x for x being greater than or equal to zero, and less than or equal to 360.
    3cos(x) + sin(x) - 1

    How would you go about solving this? I tried graphing the equation and finding the x-intercept, but the values i got didn't work for whatever reason...
     
  2. jcsd
  3. Jul 1, 2005 #2

    mathman

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    Science Advisor

    When you say 360, do you mean degrees?
    Since cos(x)=sqrt(1-sin2(x)), you can set it up as a quadratic equation in cos(x), solve and go from there - discarding any solutions where sin2+cos2 does not add up to 1.
     
  4. Jul 1, 2005 #3
    The best way when you have a linear trig. equation is to use parametric formulae

    t=tan(x/2)

    then sin(x)=2t/(t^2+1) and cos(x)=(1-t^2)/(1+t^2)

    you substitute them into the equation and solve it into t, simple, isn't it?

    Never heard about that? Quite strange.
     
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